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Eddi Din [679]
3 years ago
5

The function f(x)=3x^3+x^2+2x rises as x grows very large. A. True B. False

Mathematics
1 answer:
Alja [10]3 years ago
4 0
True I’m pretty sure
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2y to the 2nd power +11y +14
goldfiish [28.3K]

2y^2 +11y+14 =(2y+7)(y+2)

Step-by-step explanation:

Given

2y^2 +11y+14

=2y^2 +(7+4)y+14

=2y^2+7y+4y+14

= y(2y+7) + 2(2y+7)

=(2y+7)(y+2)

3 0
3 years ago
Solve using elimination.<br> 3x − 5y = 18 <br> -10x + 5y = 10
olga55 [171]

3x - 5y = 18 over

-10x + 5y = 10

Add them because the 5y's have the right symbols for us to add

-7x = 28

Divide

x = -4

Now you can plug in -4 for x in one equation, I would use the first equation!

3(-4) - 5y = 18

-12 - 5y = 18

Add 12

-5y = 30

Divide

y = -6

Your solutions are going to be:

x = -4

y = -6

To check your work plug x and y into one equation:

3(-4) - 5(-6) = 18

-12 + 30 = 18

18 = 18

Since 18 does equal 18 you know that your solution's work!

5 0
3 years ago
Read 2 more answers
Which of these values for p and a will cause the function f(x)=Pax to be an exponential growth function
vovikov84 [41]

The option are missing in the question. The options are :

A.  P = 2, a = 1

B.  $P=\frac{1}{2} ;  a =\frac{1}{3}$

C. $P=\frac{1}{2} ;  a =1$

D. P = 2, a = 3

Solution :

The given function is $f(x)= Pa^x$

So for the function to be an exponential growth, a should be a positive number and should be larger than 1. If it less than 1 or a fraction, then it is a decay. If the value of a is negative, then it would be between positive and negative alternately.

When the four option being substituted in the function, we get

A). It is a constant function since $2(1^x)=2$

B). Here, the value of a is a fraction which is less than 1, so it is a decay function. $f(x)=\frac{1}{2}\left(\frac{1}{3}\right)^x$

C). It is a constant function since the value of a is 1.

D). Here a = 3. So substituting, as the value of x increases by 1, the value of the function, f(x) increases by 3 times.

  $f(x)=2(3)^x$

Therefore, option (D). represents an exponential function.

4 0
2 years ago
What are the coordinates of the pre-image of N' if it was reflected across the y-axis, and N' has coordinates (3, -1)? Please ex
borishaifa [10]
The answer would be (3,-1) because it is reflected
6 0
3 years ago
BRAINLIESTT ASAP! PLEASE HELP ME :)
sukhopar [10]

Answer: A) none of the equations are identities

==========================================

Part 1

Plug in theta = 0

sin(theta+pi/2) - cos(theta+pi/6) = 2*cos(theta) - sin(theta)

sin(0+pi/2) - cos(0+pi/6) = 2*cos(0) - sin(0)

1 - sqrt(3)/2 = 2*1 - 0

0.13 = 2

which is a false equation

So we do not have an identity in equation 1.

-------------------------------------------

Part 2

Plug in theta = 0

sin(theta+pi/6) + cos(theta+pi/3) = (sqrt(2)/3)*sin(theta) + 2*cos(theta)

sin(0+pi/6) + cos(0+pi/3) = (sqrt(2)/3)*sin(0) + 2*cos(0)

1/2 + 1/2 = 0 + 2

1 = 2

which is also false

This is not an identity either.

6 0
2 years ago
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